Three smallest squares are chosen randomly on a chess board are the probability that these squares have exactly two corners, but no side common is:
Step-by-Step Solution
Key Concept: Three squares on a chessboard have exactly two corners but no side common when they form a diagonal pattern - specifically when two squares share only a corner vertex (diagonal adjacency) while the third square is positioned such that no two squares share an edge. This requires counting configurations where squares are at relative positions like (0,0), (1,1), and another square that maintains the constraint, using combinatorial selection from the 64 squares.
If a shaded square is chosen, 2 additional corner squares must be selected from 4 remaining corner squares, which can be done in $^4C_2 = 6$ ways. The probability is $ rac{^4C_1 \cdot 6^2}{^{64}C_3}$.
Correct Answer: 3